• HDU-5340 Three Palindromes(字符串哈希)


    http://acm.hdu.edu.cn/showproblem.php?pid=5340

    orz到了新的字符串hash姿势

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<cmath>
    #define rep(i,l,r) for (int i=l;i<=r;i++)
    #define maxn 200500
    #define p 1000000007
    using namespace std;
    typedef unsigned long long ll;
    ll a[maxn],b[maxn],k[maxn];
    int c[maxn],d[maxn],t1,t2,cnt,n,t;
    char s[maxn];
    int main(){
    scanf("%d",&t);
    k[0]=1;
    rep(i,1,20000) k[i]=k[i-1]*p;
    while (t--){
    scanf("%s",s+1);
    n=strlen(s+1);
    t1=t2=cnt=b[n+1]=0;
    for (int i=1;i<=n;i++) a[i]=a[i-1]*p+s[i];
    for (int i=n;i>=1;i--) b[i]=b[i+1]*p+s[i];
    for (int i=1;i<=n;i++){
    ll h=b[1]-b[i+1]*k[i];
    if (h==a[i]) c[++t1]=i;
    h=a[n]-a[n-i]*k[i];
    if (h==b[n-i+1]) d[++t2]=n-i+1;
    }
    reverse(c+1,c+1+t1);
    int l,r,flag=0;
    rep(i,1,t1){
    if (flag) break;
    rep(j,1,t2){
    l=c[i]+1; r=d[j]-1;
    if (l>r||cnt>2e6) break;
    ++cnt;
    ll h1=a[r]-a[l-1]*k[r-l+1];
    ll h2=b[l]-b[r+1]*k[r-l+1];
    if (h1==h2) flag=1;
    }
    }
    if (flag) printf("Yes ");
    else printf("No ");
    }
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/ctlchild/p/4694871.html
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